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The numbers D, N, and P are positive integers, such that D < N, and N

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The numbers D, N, and P are positive integers, such that D < N, and N  [#permalink]

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New post 17 Mar 2015, 05:58
6
29
00:00
A
B
C
D
E

Difficulty:

  95% (hard)

Question Stats:

25% (02:58) correct 75% (02:31) wrong based on 388 sessions

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The numbers D, N, and P are positive integers, such that D < N, and N  [#permalink]

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New post 23 Mar 2015, 05:57
2
8
Bunuel wrote:
The numbers D, N, and P are positive integers, such that D < N, and N is not a power of D. Is D a prime number?

(1) N has exactly four factors, and D is a factor of N
(2) D = (3^P) + 2


Kudos for a correct solution.


MAGOOSH OFFICIAL SOLUTION:

This is a tricky one.

Statement #1: two kinds of numbers have exactly four factors: (a) products of two distinct prime numbers, and (b) cubes of prime numbers.

The product of two distinct prime numbers S and T would have factors {1, S, T, ST}. For example, the factors of 10 are (1, 2, 5, 10), and the factors of 21 are {1, 3, 7, 21}.

The cube of a prime number S would have as factors 1, S, S squared, and S cubed. For example, 8 has factors {1, 2, 4, 8} and 27 has factors {1, 3, 9, 27}.

We know N is not a power of D, so the second case is excluded. N must be the product of two distinct prime numbers. We know D < N, so of the four factors, D can’t be the product of the two prime numbers. D could be either of the prime number factors, or D could be 1, which is not a prime number. Because D could either be a prime number or 1, we cannot give a definitive answer to the question. This statement, alone and by itself, is not sufficient.

Statement #2: this is tricky. The first few plug-ins seem to reveal a pattern.

Image

Even if you sense a pattern, it’s important to remember that plugging in numbers alone is never enough to establish that a DS statement is sufficient. Here, if we persevered to one more plug-in, we would find the one that breaks the pattern.
P = 5 --> 3^5 + 2 = 243, which is not a prime.

That gives another answer to the prompt, so we know this statement is not sufficient.

To avoid a lot of plugging in, it’s also very good to know that in mathematics, prime numbers are notorious for not following any easy pattern. It is impossible to produce an algebraic formula that will always produce prime numbers. In fact, this is more than you need to know, but the hardest unsolved question in higher mathematics, the Riemann Hypothesis, concerns the pattern of prime numbers; mathematicians have been working on this since 1859, and no one has proven it yet. Suffice to say that no one-line algebraic formula is going to unlock the mystery of prime numbers!

Combined statements: according to the information in statement #1, either D = 1 or D is a prime number. Well, statement #2 excludes the possibility that D = 1, because that number cannot be written as two more than a power of 3. Therefore, D must be a prime number. We have a definitive answer to the prompt question. Combined, the statements are sufficient.

Answer = (C)

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Re: The numbers D, N, and P are positive integers, such that D < N, and N  [#permalink]

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New post 17 Mar 2015, 07:51
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I will say C. Please let me know if I am missing something.

(1) The four factors of N include 1, D, something, and N. Therefore, N is the product of D and something.
Test case 1: N = 6 so D is either 2 or 3 and therefore is prime.
Test case 2: N = 8 so D is 4 and not prime (D cannot be 2 because 8 is a power of 2, which violates the constraints.)


(2) Possible values for D include 5, 11, 29, 83, 245, ... etc.
We have prime and non-prime possibilities, so this is insufficient.

Combining the statements, D must be one of the prime values given in statement 2, otherwise N would have more than four factors.
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Re: The numbers D, N, and P are positive integers, such that D < N, and N  [#permalink]

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New post 17 Mar 2015, 08:10
Bunuel wrote:
The numbers D, N, and P are positive integers, such that D < N, and N is not a power of D. Is D a prime number?

(1) N has exactly four factors, and D is a factor of N
(2) D = (3^P) + 2


Kudos for a correct solution.



Option A:N has exactly four factors, and D is a factor of N : Not Sufficient
\(2^1 * 3^1\) factors are 1,2,3,6 ; D can be 2 or 1 . Not sufficient.
\(19^1 * 2^1\) factors are 1,2,19,38 ; D can be 19,2,1 . (note 1 is not prime)
Option B: D = (3^P) + 2 Not Sufficient
\(P=1; D=5 is prime ? Yes\)
\(P=5; D=245 is prime ? NO\)

A && B
from B we know that D is and ODD Integer and from A we have
D can be 19,2,1 . (note 1 is not prime) 19 is prime but 1 is not so still the statement is insufficient.

Answer E
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The numbers D, N, and P are positive integers, such that D < N, and N  [#permalink]

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New post 17 Mar 2015, 10:02
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Lucky2783 wrote:
D can be 19,2,1 . (note 1 is not prime) 19 is prime but 1 is not so still the statement is insufficient.


If D = 1, then 3^P = -1 and P is not a positive integer, violating the constraint given in the problem. (Same with D = 2 or 19)
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Re: The numbers D, N, and P are positive integers, such that D < N, and N  [#permalink]

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New post 17 Mar 2015, 10:24
sterling19 wrote:
Lucky2783 wrote:
D can be 19,2,1 . (note 1 is not prime) 19 is prime but 1 is not so still the statement is insufficient.


If D = 1, then 3^P = -1 and P is not a positive integer, violating the constraint given in the problem. (Same with D = 2 or 19)



good catch !!
Answer is C.
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Re: The numbers D, N, and P are positive integers, such that D < N, and N  [#permalink]

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New post 23 Mar 2015, 06:50
Lucky2783 wrote:
sterling19 wrote:
Lucky2783 wrote:
D can be 19,2,1 . (note 1 is not prime) 19 is prime but 1 is not so still the statement is insufficient.


If D = 1, then 3^P = -1 and P is not a positive integer, violating the constraint given in the problem. (Same with D = 2 or 19)



good catch !!
Answer is C.



Hi Bunuel,

We (me and @sterling19) have come up with number substitution solution for this question but it took me 2-3 mins to write the test cases ,
is there a better way to attack this question apart from the magoosh official solution ?
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The numbers D, N, and P are positive integers, such that D < N, and N  [#permalink]

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New post 17 Nov 2015, 08:58
Bunuel
Bunuel wrote:
Bunuel wrote:
The numbers D, N, and P are positive integers, such that D < N, and N is not a power of D. Is D a prime number?

(1) N has exactly four factors, and D is a factor of N
(2) D = (3^P) + 2


Kudos for a correct solution.


MAGOOSH OFFICIAL SOLUTION:

This is a tricky one.

Statement #1: two kinds of numbers have exactly four factors: (a) products of two distinct prime numbers, and (b) cubes of prime numbers.

The product of two distinct prime numbers S and T would have factors {1, S, T, ST}. For example, the factors of 10 are (1, 2, 5, 10), and the factors of 21 are {1, 3, 7, 21}.

The cube of a prime number S would have as factors 1, S, S squared, and S cubed. For example, 8 has factors {1, 2, 4, 8} and 27 has factors {1, 3, 9, 27}.

We know N is not a power of D, so the second case is excluded. N must be the product of two distinct prime numbers. We know D < N, so of the four factors, D can’t be the product of the two prime numbers. D could be either of the prime number factors, or D could be 1, which is not a prime number. Because D could either be a prime number or 1, we cannot give a definitive answer to the question. This statement, alone and by itself, is not sufficient.

Statement #2: this is tricky. The first few plug-ins seem to reveal a pattern.

Image

Even if you sense a pattern, it’s important to remember that plugging in numbers alone is never enough to establish that a DS statement is sufficient. Here, if we persevered to one more plug-in, we would find the one that breaks the pattern.
P = 5 --> 3^5 + 2 = 243, which is not a prime.

That gives another answer to the prompt, so we know this statement is not sufficient.

To avoid a lot of plugging in, it’s also very good to know that in mathematics, prime numbers are notorious for not following any easy pattern. It is impossible to produce an algebraic formula that will always produce prime numbers. In fact, this is more than you need to know, but the hardest unsolved question in higher mathematics, the Riemann Hypothesis, concerns the pattern of prime numbers; mathematicians have been working on this since 1859, and no one has proven it yet. Suffice to say that no one-line algebraic formula is going to unlock the mystery of prime numbers!

Combined statements: according to the information in statement #1, either D = 1 or D is a prime number. Well, statement #2 excludes the possibility that D = 1, because that number cannot be written as two more than a power of 3. Therefore, D must be a prime number. We have a definitive answer to the prompt question. Combined, the statements are sufficient.

Answer = (C)


if N is a cube of a prime, n = 27
in this case we can have d as either 1 or 9, right?
why have we completely ignored the case of n being a cube of a prime no?
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Re: The numbers D, N, and P are positive integers, such that D < N, and N  [#permalink]

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Re: The numbers D, N, and P are positive integers, such that D < N, and N   [#permalink] 29 Jan 2019, 04:41
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